solutions and answers needed to question 3 (3) A student buys a brand new probab
ID: 3067833 • Letter: S
Question
solutions and answers needed to question 3
(3) A student buys a brand new probability textbook that has 500 pages, num- bered from 001 to 500. She randomly opens the book to a page and starts to read. Assume that any of the 500 pages are equally likely to be chosen. Using inclusion and exclusion to find the probability that the page num- ber she chooses contains at least one "5" as a digit. (4) Keeping the set-up in Question (3). What is the probability that the page number contains at least on "3" as a digit,Explanation / Answer
(3)
Let B be the event that at least one of the digits on the chosen page is a "5".
5 can be on three places. ones, tens and hundreds.
Excluding 500th page:
Number of ways when 5 is at ones places( _,_, 5) = 5 × 10 = 50
Number of ways when 5 is at tens place (_,5,_) = 5 x 10 = 50
Number of ways when 5 is at hundreds place (5,_,_) = 0 (Since we excluded 500th page)
N(B) = 50 + 50 + 0
N(B) = 100
Total number of pages = N(S) = 499 (Since we excluded 500th page)
P(B) = 100/499 = 0.200
Including 500th page:
Number of ways when 5 is at ones places( _,_, 5) = 5 × 10 = 50
Number of ways when 5 is at tens place (_,5,_) = 5 x 10 = 50
Number of ways when 5 is at hundreds place (5,_,_) = 1 x 1 =1 (Since we included 500th page and here we use only one number that is 0 in the two places)
N(B) = 50 + 50 + 1
N(B) = 101
Total number of pages = N(S) = 500 (Since we included 500th page)
P(B) = 101/500 = 0.202
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